Question

Find the value of the trigonometric functions for the following:
sec θ = `13/5`, θ lies in the IV quadrant

Answer 1

We know that sec²θ – tan²θ = 1

`(13/2)^2 - tan^2theta` = 1

`169/25 - 1` = tan²θ

tan²θ = `(169 - 25)/25 = 144/25`

tan θ = `+-  12/5`

Since θ lies in the fourth quadrant tan θ is negative.

∴ tan θ = `- 12/5`

cos θ = `1/sectheta = 5/13`

We know sin²θ + cos²θ = 1

`sin^2theta + (5/13)^2` = 1

sin²θ = `1 - 25/169`

sin²θ = `(169 - 25)/169 = 144/169`

sin θ = `+- 12/13`

Since θ lies in the fourth quadrant sin θ is negative.

∴ sin θ = `- 12/13`

sin θ = `- 12/13`, cosec θ = `1/sintheta = - 13/12`

cos θ = `5/13`, sec θ = `1/costheta = 13/5`

tan θ = `- 12/5`, cot θ = `1/tantheta = - 5/12`